Answer
$\dfrac{1}{6}$
Work Step by Step
Factoring the expressions and then cancelling the common factor/s between the numerator and the denominator, the given expression, $
\dfrac{x^2-x-12}{2x^2-32}\cdot \dfrac{x^2+8x+16}{3x^2+21x+36}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{(x-4)(x+3)}{2(x^2-16)}\cdot \dfrac{(x+4)(x+4)}{3(x^2+7x+12)}
\\\\=
\dfrac{(x-4)(x+3)}{2(x+4)(x-4)}\cdot \dfrac{(x+4)(x+4)}{3(x+4)(x+3)}
\\\\=
\dfrac{(\cancel{x-4})(\cancel{x+3})}{2(\cancel{x+4})(\cancel{x-4})}\cdot \dfrac{(\cancel{x+4})(\cancel{x+4})}{3(\cancel{x+4})(\cancel{x+3})}
\\\\=
\dfrac{1}{6}
.\end{array}
